Proving a fact about inner product spaces

[Mr Davis 97]

Homework Statement

Let $V$ be a vector space equipped with an inner product $\langle \cdot, \cdot \rangle$. If $\langle x,y \rangle = \langle x, z\rangle$ for all $x \in V$, then $y=z$.

Homework Equations

The Attempt at a Solution

Here is my attempt. $\langle x,y \rangle = \langle x, z\rangle$ means that $\langle x,y \rangle - \langle x, z\rangle = 0$, and so $\langle x, y - z\rangle = 0$. This is as far as I've gotten. I don't see how to deduce that $y - z = 0$.

 

[fresh_42]

Homework Statement

Let $V$ be a vector space equipped with an inner product $\langle \cdot, \cdot \rangle$. If $\langle x,y \rangle = \langle x, z\rangle$ for all $x \in V$, then $y=z$.

Homework Equations

The Attempt at a Solution

Here is my attempt. $\langle x,y \rangle = \langle x, z\rangle$ means that $\langle x,y \rangle - \langle x, z\rangle = 0$, and so $\langle x, y - z\rangle = 0$. This is as far as I've gotten. I don't see how to deduce that $y - z = 0$.

It's hidden in the term "inner product". In the sense used here, it is non-degenerated, which means exactly this: $\langle x,y \rangle=0 \; \forall \; x \in V \Longrightarrow y=0$. This is true for the usual dot product like in real vector spaces. It is not automatically true for an arbitrary bilinear form. So in concrete cases, one has to check the properties of the "inner product" used.

 

[Mr Davis 97]

It's hidden in the term "inner product". In the sense used here, it is non-degenerated, which means exactly this: $\langle x,y \rangle=0 \; \forall \; x \in V \Longrightarrow y=0$. This is true for the usual dot product like in real vector spaces. It is not automatically true for an arbitrary bilinear form. So in concrete cases, one has to check the properties of the "inner product" used.

So are you saying that $\langle x,y \rangle=0 \; \forall \; x \in V \Longrightarrow y=0$ is a property of inner product spaces in general?

 

[fresh_42]

So are you saying that $\langle x,y \rangle=0 \; \forall \; x \in V \Longrightarrow y=0$ is a property of inner product spaces in general?

No, I'm not saying this, because I'm used to deal with bilinear forms, that do not obey this rule and I don't see why they shouldn't be called inner product, too. So I am saying: look at your definition. This is even more important in the complex case, in which the product isn't symmetric anymore. (See https://en.wikipedia.org/wiki/Inner_product_space#Alternative_definitions.2C_notations_and_remarks.)

The convention however is, that inner products have to be positive definite, that is $\langle x,y \rangle \geq 0$ and $\langle x,x \rangle=0 \Longleftrightarrow x=0$. With this convention we get in your case $\forall_{x}\; \langle x,y-z \rangle=0 \Longrightarrow \langle y-z,y-z \rangle=0 \Longrightarrow y-z=0$.

 

[Mr Davis 97]

No, I'm not saying this, because I'm used to deal with bilinear forms, that do not obey this rule and I don't see why they shouldn't be called inner product, too. So I am saying: look at your definition. This is even more important in the complex case, in which the product isn't symmetric anymore. (See https://en.wikipedia.org/wiki/Inner_product_space#Alternative_definitions.2C_notations_and_remarks.)

The convention however is, that inner products have to be positive definite, that is $\langle x,y \rangle \geq 0$ and $\langle x,x \rangle=0 \Longleftrightarrow x=0$. With this convention we get in your case $\forall_{x}\; \langle x,y-z \rangle=0 \Longrightarrow \langle y-z,y-z \rangle=0 \Longrightarrow y-z=0$.

Could you explain the implication $\forall_{x}\; \langle x,y-z \rangle=0 \Longrightarrow \langle y-z,y-z \rangle=0$?

 

[fresh_42]

Could you explain the implication $\forall_{x}\; \langle x,y-z \rangle=0 \Longrightarrow \langle y-z,y-z \rangle=0$?

Yes.

If $\langle x,y \rangle = \langle x, z\rangle$ for all $x\in V$

... so $\langle x, y - z\rangle = 0$.

... still for all $x\in V$, so especially for $x:=y-z$.

 

[nuuskur]

As fresh_42 said, it holds for all $x$ including $x:=y-z$. There is an axiom of scalar product, that gives you the answer of what $y-z$ can be if its length is zero.

 

[member 587159]

No, I'm not saying this, because I'm used to deal with bilinear forms, that do not obey this rule and I don't see why they shouldn't be called inner product, too. So I am saying: look at your definition. This is even more important in the complex case, in which the product isn't symmetric anymore. (See https://en.wikipedia.org/wiki/Inner_product_space#Alternative_definitions.2C_notations_and_remarks.)

The convention however is, that inner products have to be positive definite, that is $\langle x,y \rangle \geq 0$ and $\langle x,x \rangle=0 \Longleftrightarrow x=0$. With this convention we get in your case $\forall_{x}\; \langle x,y-z \rangle=0 \Longrightarrow \langle y-z,y-z \rangle=0 \Longrightarrow y-z=0$.

Don't you mean $<x,x> \geq 0$?

 

[fresh_42]

Don't you mean $<x,x> \geq 0$?

Yes, of course. Copy and paste error.

 

[WWGD]

Specifically, positive definiteness https://en.wikipedia.org/wiki/Inner_product_space